POJ 2386 Lake Counting（水淹菜地，DFS，八连通，连通分量）

本文链接：https://blog.youkuaiyun.com/shyazhut/article/details/52056546

本文介绍了一个湖泊计数问题，通过DFS算法找出地图上由字母'W'表示的水域形成的独立湖泊数量。题目来源于USACO2004年十一月赛，提供了完整的代码实现。

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29064		Accepted: 14543

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

Source

USACO 2004 November

题意：

让我们求出被水淹没的区域有几块。（字母 ’ W ‘ 表示水淹没的区域）

思路：

简单的DFS，求出八连通的连通分量就行了。

代码：

#include<stdio.h>
#include<string.h>
#define MYDD 1103

int N,M;
char map[128][128];//记录积水的坐标点
int dx[]= {0,0,-1,1,1,1,-1,-1};//八连通的图
int dy[]= {-1,1,0,0,-1,1,-1,1};//当前位置周围的八个方向

void DFS(int x,int y) {
	map[x][y]='.';//置当前位置没有积水
	for(int j=0; j<8; j++) {
		int gx=x+dx[j];
		int gy=y+dy[j];
		if(gx>=0&&gx<=N&&gy>=0&&gy<=M&&map[gx][gy]=='W') {
			DFS(gx,gy);//满足条件继续查询
		}
	}
	return ;
}

int main() {
	scanf("%d%d",&N,&M);
	for(int j=0; j<N; j++)
		scanf("%s",map[j]);//键入地图
	int ans=0;//记录答案
	for(int j=0; j<N; j++) {
		for(int k=0; k<M; k++) {
			if(map[j][k]=='W') {//从有积水的地方遍历
				DFS(j,k);
				ans++;
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}